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sveta [45]
3 years ago
7

You have a mixture that contains 0.380 moles of Ne(g), 0.250 moles of He(g), and 0.500 moles CH4(g) at 400 K and 6.25 atm. What

is the partial pressure of He?
Chemistry
1 answer:
Bas_tet [7]3 years ago
4 0

Answer:

1.38 atm

Explanation:

Partial pressure of a gas = total pressure × mole fraction of that gas.

Mole fraction of a gas is moles of a gas ÷ total moles of all gases.

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An aqueous potassium carbonate solution is made by dissolving 5.51 moles of K 2 CO 3 in sufficient water so that the final volum
ad-work [718]

Answer:

1.67mol/L

Explanation:

Data obtained from the question include:

Mole of solute (K2CO3) = 5.51 moles

Volume of solution = 3.30 L

Molarity =?

Molarity is simply the mole of solute per unit litre of the solution. It can be expressed mathematically as:

Molarity = mole of solute /Volume of solution

Molarity = 5.51 mol/3.30 L

Molarity = 1.67mol/L

Therefore, the molarity of K2CO3 is 1.67mol/L

3 0
3 years ago
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Jim ran 200 meters in 30.4 seconds. Mac ran 100 meters in 30.4 seconds. Who ran faster?
VladimirAG [237]

Jim, because he ran a greater distance in the same time :)

By the way, this is a maths question

7 0
4 years ago
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What is the theoretical yield if 35.5g of Al reacts 39.0g of Cl2
atroni [7]

Answer : The correct answer for the Theoretical Yield is 48.93 g of product .

Theoretical yield : It is amount of product produced by limiting reagent . It is smallest product yield of product formed .

Following are the steps to find theoretical yield .

Step 1) : Write a balanced reaction between Al and Cl₂ .

2 Al + 3 Cl₂→ 2 AlCl₃

Step 2: To find amount of product (AlCl₃) formed by Al .

Following are the sub steps to calculate amount of AlCl₃ formed :

a) To calculate mole of Al :

Given : Mass of Al = 35.5 g

Mole can be calculate by following formula :

Mole = \frac{given mass (g)}{atomic mass \frac{g}{mol}}

Mole = \frac{35.5 g }{26.9 \frac{g}{mol}}

Mole = 1.32 mol

b) To find mole ratio of AlCl₃ : Al

Mole ratio is calculated from balanced reaction .

Mole of Al in balanced reaction = 2

Mole of AlCl₃ in balanced reaction = 2.

Hence mole ratio of AlC; l₃ : Al = 2:2

c) To find mole of AlCl₃ formed :

Mole of AlCl_3 = Mole of Al * Mole ratio

Mole of AlCl_3 = 1.32 mol of Al * \frac{2}{2}

Mole of AlCl₃ = 1.32 mol

d) To find mass of AlCl₃

Molar mass of AlCl₃ = 133.34 \frac{g}{mol}

Mass of AlCl3 can be calculated using mole formula as:

1.32 mol of AlCl_3 = \frac{ mass (g)}{133.34 \frac{g}{mol}}

Multiplying both side by 133.34 \frac{g}{mol}

1.32 mole  * 133.34\frac{g}{mol} = \frac{mass (g)}{133.34\frac{g}{mol}} *133.34  \frac{g}{mol}

Mass of AlCl₃ = 176.00 g

Hence mass of AlCl₃ produced by Al is 176.00 g

Step 3) To find mass of product (AlCl₃) formed by Cl₂ :

Same steps will be followed to calculate mass of AlCl₃

a) Find mole of Cl₂

Mole of Cl_2 = \frac{39.0 g}{70.9\frac{g}{mol}}

Mole of Cl₂ = 0.55 mol

b) Mole ratio of Cl₂ : AlCl₃

Mole of Cl₂ in balanced reaction = 3

Mole of AlCl₃ in balanced reaction = 2

Hence mole ratio of AlCl₃ : Cl₂ = 2 : 3

c) To find mole of AlCl₃

Mole of AlCl_3 = mole of Cl_2 * mole ratio

Mole of AlCl_3 = 0.55  mole  * \frac{2}{3}

Mole of AlCl3 = 0.367 mol

d) To find mass of AlCl₃ :

0.367 mol of AlCl_3 = \frac{mass (g) }{133.34 \frac{g}{mol}}

Multiplying both side by

133.34 \frac{g}{mol}

0.367 mol of AlCl_3 * 133.34 \frac{g}{mol}  = \frac{mass(g)}{133.34\frac{g}{mol}}   * 133.34 \frac{g}{mol}

Mass of AlCl₃ = 48.93 g

Hence mass of AlCl₃ produced by Cl₂ = 48.93 g

Step 4) To identify limiting reagent and theoretical yield :

Limiting reagent is the reactant which is totally consumed when the reaction is complete . It is identified as the reactant which produces least yield or theoretical yield of product .

The product AlCl₃ formed by Al = 176.00 g

The product AlCl₃ formed by Cl₂ = 48.93 g

Since Cl₂ is producing less amount of product hence it is limiting reagent and 48.93 g will be considered as Theoretical yield .

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3 years ago
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Two reactants, each having a mass of 50.0 g, react fully in a beaker. What is the total mass of the products in the reaction?
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6 0
2 years ago
Consider the following reversible reaction.
dmitriy555 [2]

Answer:

Keq = [CO₂]/[O₂]

Explanation:

Step 1: Write the balanced equation for the reaction at equilibrium

C(s) + O₂(g) ⇄ CO₂(g)

Step 2: Write the expression for the equilibrium constant (Keq)

The equilibrium constant is equal to the product of the concentrations of the products raised to their stoichiometric coefficients divided by the product of the concentrations of the reactants raised to their stoichiometric coefficients. It only includes gases and aqueous species. The equilibrium constant for the given system is:

Keq = [CO₂]/[O₂]

6 0
3 years ago
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